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Svet_ta [14]
2 years ago
15

A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50

N, what is the resultant forward force on the suitcase?
Physics
1 answer:
stira [4]2 years ago
4 0
Fortunately, 'force' is a vector.  So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward.   Hold that thought, while I slog through
the complete solution of the problem.


            (100 N forward) plus (50 N backward)

        =  (100 N forward) minus (50 N forward)

        =           50 N forward .

That's it.
Is there any part of the solution that's not clear ?

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Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
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A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

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Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
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F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
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